When an ionic compound, such as a salt, is placed into water, it dissolves to some extent. Solubility describes the maximum amount of a solute that can dissolve in a solvent at a specific temperature. While many salts are highly soluble, others are sparingly soluble, meaning only a small fraction of the solid material dissolves into ions. To quantify the solubility of these sparingly soluble ionic solids, chemists use the Solubility Product Constant, or \(K_{sp}\). The \(K_{sp}\) is a measure of the equilibrium established between the undissolved solid compound and its constituent dissolved ions in a saturated solution.
Defining the Solubility Product Constant
When a sparingly soluble ionic solid is placed in water, a dynamic equilibrium is eventually reached in a saturated solution. In this state, the rate at which the solid dissolves and produces ions exactly matches the rate at which the dissolved ions recombine and precipitate back into the solid form.
The \(K_{sp}\) is a constant value for a given substance, provided the temperature remains fixed. The value of \(K_{sp}\) changes with temperature because solubility is temperature-dependent. For most ionic solids, solubility and \(K_{sp}\) tend to increase as the temperature rises.
The constant represents the product of the molar concentrations of the ions present in the solution at equilibrium. \(K_{sp}\) is calculated only when the solution is saturated, reflecting the maximum possible concentration of ions that can exist before precipitation occurs. Focusing on the dissolved ions, \(K_{sp}\) allows scientists to assess the degree to which a compound dissociates in water.
Writing the Ksp Expression
The mathematical expression for \(K_{sp}\) is derived directly from the balanced chemical equation for the dissolution of the ionic solid. Consider a general sparingly soluble salt \(A_xB_y\), which dissolves in water to produce \(x\) moles of cation \(A^{y+}\) and \(y\) moles of anion \(B^{x-}\). The balanced dissolution equation is \(A_xB_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq)\).
To write the \(K_{sp}\) expression, the concentrations of the dissolved ions are multiplied together. A fundamental rule for all equilibrium constants is that pure solids and pure liquids are excluded because their concentrations remain constant. Therefore, the solid reactant \(A_xB_y(s)\) is omitted from the final mathematical formula.
Each ion concentration is raised to the power of its stoichiometric coefficient from the balanced equation. For the general salt \(A_xB_y\), the \(K_{sp}\) expression is \(K_{sp} = [A^{y+}]^x [B^{x-}]^y\). The square brackets denote the molar concentration of the ion in moles per liter.
For example, to derive the \(K_{sp}\) expression for calcium fluoride, \(\text{CaF}_2\), the dissolution equation is \(\text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^{-}(aq)\). The resulting expression is \(K_{sp} = [\text{Ca}^{2+}][\text{F}^{-}]^2\). The concentration of the fluoride ion is squared because its stoichiometric coefficient in the balanced equation is two.
What Ksp Values Reveal About Solubility
The magnitude of the \(K_{sp}\) value provides a direct way to compare the solubilities of different ionic compounds. A substance with a large \(K_{sp}\) value is more soluble because a higher concentration of ions is required to satisfy the equilibrium condition. Conversely, a small \(K_{sp}\) value indicates that the substance is sparingly soluble, as the product of the ion concentrations remains low at equilibrium.
For instance, silver chloride (\(\text{AgCl}\)) has a \(K_{sp}\) of \(1.8 \times 10^{-10}\), while copper(I) chloride (\(\text{CuCl}\)) has a \(K_{sp}\) of \(1.2 \times 10^{-6}\). Since the \(K_{sp}\) for \(\text{AgCl}\) is smaller, it is less soluble in water than \(\text{CuCl}\). This comparison works best for compounds that dissociate into the same total number of ions, such as comparing two salts that both produce two ions upon dissolution.
The \(K_{sp}\) is conceptually linked to molar solubility, which is the number of moles of the solute that dissolve per liter of solution, often denoted by the variable \(s\). For a simple salt like \(\text{AgCl}\), \(K_{sp} = s^2\), demonstrating a direct quantitative relationship between the constant and how much of the solid dissolves.
Predicting Precipitation Using the Ion Product
The practical application of \(K_{sp}\) is predicting whether a precipitate will form when two solutions containing dissolved ions are mixed. For this prediction, scientists use the Ion Product, denoted as \(Q\). The mathematical expression for \(Q\) is identical to the \(K_{sp}\) expression, but \(Q\) uses the instantaneous concentrations of the ions, which are not necessarily equilibrium concentrations.
By comparing the calculated value of \(Q\) to the known \(K_{sp}\) for the potential precipitate, the state of the solution can be determined.
If the Ion Product (\(Q\)) is less than \(K_{sp}\), the solution is unsaturated, and no precipitate will form. When \(Q\) is equal to \(K_{sp}\), the solution is saturated, meaning the system is at equilibrium.
If \(Q\) is greater than \(K_{sp}\), the solution is supersaturated, and a precipitate will spontaneously form. The excess ions will combine to produce the solid ionic compound until the ion concentrations decrease to the point where \(Q\) is once again equal to \(K_{sp}\), establishing equilibrium.